a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)
\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{5\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b: \(x=\sqrt{17-12\sqrt{2}}+2\sqrt{2}-2\)
\(=\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}+2\sqrt{2}-2\)
\(=\sqrt{\left(3-2\sqrt{2}\right)^2}+2\sqrt{2}-2\)
\(=3-2\sqrt{2}+2\sqrt{2}-2=1\)
Thay x=1 vào A, ta được:
\(A=\dfrac{3\cdot1}{1+2}=\dfrac{3}{3}=1\)
c: Để A nguyên thì \(3\sqrt{x}⋮\sqrt{x}+2\)
=>\(3\sqrt{x}+6-6⋮\sqrt{x}+2\)
=>\(-6⋮\sqrt{x}+2\)
=>\(\sqrt{x}+2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
=>\(\sqrt{x}\in\left\{-1;-3;0;-4;1;-5;4;-8\right\}\)
=>\(\sqrt{x}\in\left\{0;4;1\right\}\)
=>\(x\in\left\{0;16;1\right\}\)
